unit 7 (chp 14): chemical kinetics (rates) john d. bookstaver st. charles community college st....
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Unit 7 (Chp 14):
Chemical Kinetics(rates)
John D. Bookstaver
St. Charles Community College
St. Peters, MO
2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chemical Kinetics is the study of:
1) Rates at which reactants are consumed or products are produced in a chemical rxn.
2) Factors that affect the rate of a reaction according to Collision Theory (temperature, concentration, surface area, & catalyst).
3) Mechanisms, or sequences of steps, for how a reaction actually occurs.
4) Rate Laws (equations) used to calculate rates, rate constants, concentrations, & time.
Chemical Kinetics
Reaction Rates
Reaction rates are described by the change in concentration (DM) of reactants (consumed) or of products (produced) per change in time.
Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
[C4H9Cl] measured at various times
recall:[ ] brackets represent concentration in molarity (M)
Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
[C4H9Cl]t
“change in”
Reaction Rates
fewer reactant collisions
average rate slows
WHY?
C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
• The slope of a line tangent to the curve at any point is the instantaneous rate at that time.
Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
[C4H9Cl]t
(rise)(run)
• All reactions slow down over time.
• The initial rate of reaction is commonly chosen for analysis and comparison.
Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
rate of consumptionof reactant
=rate of production
of product.
(IF 1:1 mol ratio)
–[C4H9Cl]t
= [C4H9OH]t
↓ reactant = ↑ product
Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq)
Rate:
Reaction Rates:
3. production of products per time
2. consumption of reactants per time
4. Are equalized to a stoich. coefficient of 1
2 NO
1 O2
2 NO2 2 NO + O2
–[NO2]2 t
=[NO]2 t =
[O2]t
–[NO2]t
+[NO]t
[NO2]t
1. slope =
5. change during rxn time (t)
change in conc.change in time
conc
entr
atio
n (M
)
Initial rate of production of H2 is 0.050 M∙s–1.What is the rate of consumption of HI?
aA + bB cC + dD
Rate Ratios
Rate 1a
[A]t =
1b
[B]t =
1c
[C]t
1d
[D]t=
2 HI(g) H2(g) + I2(g)
− −=
0.050 M∙s–1 H2 xmol H2
mol HI1
=2 0.10 M∙s–1 HI
–0.10 M∙s–1 HImolL∙s mol ratio
=?
HW p. 619 #20
Rate equations (or rate laws) have the form:
Rate Laws
rate = k[A]x[B]y[C]z
rate constant
order with respect to reactantsA, B, & C
…or…number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
rate = k[BrO3–][Br–][H+]2
Example: overall order= ___ order4th
(4 particles in collision)
• reactant bonds break, then product bonds form.• Reaction rates depend on collisions
between reactant particles with:
1) greater frequency
2) enough energy
3) proper orientation
The Collision Model
reactants productsCollision
activation energy:minimum E required
to start reaction(Eact )
successful
unsuccessful
Reaction Coordinate Diagramtransition state
∆E
Eact
…aka…EnergyProfile
Pot
entia
l Ene
rgy
Reaction progress
demo
4 Factors that Affect Reaction Rates:
1) Concentration
2) Temperature
3) (exposed) Surface Area (particle size)
4) Catalyst
Factors That Affect Reaction Rates
1) Concentration of Reactants ↑ concentration, ↑ collision frequency increase pressure of gases
Factors That Affect Reaction Rates
Fe(s) + O2(g) Fe2O3(s) 20% of air
is O2(g)
100% O2(g)
Eact at higher Temp more particles
over Eact
collisions of…greater frequencygreater energy
unsuccessful collisions(bounce off)
Factors That Affect Reaction Rates2) Temperature
successful collisions(react)
↑ Temp, ↑ rate
↑ Temp, ↑ ratek is temp. dependent
(k changes with temp)
Temperature and k (rate constant)
rate = k [A]x
3) Surface Area (particle size) smaller pieces, more exposed
surface area for collision.
Factors That Affect Reaction Rates
consumed, then produced (not used up)
4) Catalyst
Factors That Affect Reaction Rates
Uncatalyzed …lowering the Eact .
reaction progress
pote
ntia
l ene
rgy
2 H2O2 2 H2O + O2
2 H2O2
2 H2O + O2
+ 2 Br– + 2 H+
+ 2 Br– + 2 H+
Br2(intermediate)
Catalyzed
↑ rate by changing the reaction mechanism by…
Br–, H+
demo
Surface CatalystsCatalysts can orient reactants to help bonds break and form.
H2 + H2C=CH2 H3C–CH3
H2 + H2C=CH2
CH3CH3
• biological catalysts in living systems.• A substrate fits into the active site of the
enzyme much like a key fits into a lock.
(IMAFs work here)
Enzymes
HW p. 621#50,51,64
Rate equations (or rate laws) have the form:
Rate Laws
rate = k[A]x[B]y[C]z
rate constant
order with respect to reactantsA, B, & C
…or…number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
rate = k[BrO3–][Br–][H+]2
Example: overall order= ___ order4th
(4 particles in collision)
recall…
Reaction MechanismsThe sequence of molecular collisions and changes by which reactants become products is called the reaction mechanism.
• Rxns may occur in separate elementary steps.
• The overall reaction occurs only as fast as the slowest, rate-determining step. (RDS)
Slow First Step
• A proposed mechanism for this reaction is:
Step 1: NO2 + NO2 NO3 + NO (slow)
Step 2: NO3 + CO NO2 + CO2 (fast)
• NO3 intermediate is produced then consumed.
NO2 (g) + CO (g) NO (g) + CO2 (g)
• CO is not involved in the slow RDS,
so it does not appear in the rate law.
(OR…the order w.r.t. CO is ___)
• Rate law depends on the slow 1st step:
rate = k [NO2]2
0th
Slow Second Step
• A proposed mechanism is:
Step 2: NOBr2 + NO 2 NOBr (slow)
Step 1: NO + Br2 NOBr2 (fast)
2 NO(g) + Br2(g) 2 NOBr(g)
• NOBr2 intermediate is produced then consumed.
• Rate law depends on the slow 2nd step:
rate = k2 [NOBr2] [NO]
• But cannot include intermediate [NOBr2] (b/c
it’s difficult to control conc.’s of intermediates).
Slow Second Step
b/c step 1 is in equilibrium
Step 2: NOBr2 + NO 2 NOBr (slow)
Step 1: NO + Br2 NOBr2 (fast)
From Step 1: rateforward = ratereverse
k1 [NO] [Br2] = k−1 [NOBr2]
k1
k−1
[NO] [Br2] = [NOBr2]
From RDS Step 2: rate = k2 [NOBr2] [NO]
solve for [NOBr2]
substitute for [NOBr2]
Slow Second Step
Step 2: NOBr2 + NO 2 NOBr (slow)
Step 1: NO + Br2 NOBr2 (fast)
k1
k−1
[NO] [Br2] = [NOBr2]
rate = k2 [NOBr2] [NO]
substitute for [NOBr2]
k2k1
k−1
rate = [NO] [Br2] [NO]
rate = k [NO]2 [Br2]
HW p. 625 #92, 94
WS Reaction Mechanisms
Rate equations (or rate laws) have the form:
Rate Laws
rate = k[A]x[B]y[C]z
rate constant
order with respect to reactantsA, B, & C
…or…number of each particle involved
in collision that affects the rate
overall order of reaction = x + y +…
rate = k[BrO3–][Br–][H+]2
Example: overall order= ___ order4th
(4 particles in collision)
recall…
Orders from Experimental Data: Rate varies with Concentration
Initial Concentrations Rate in M per unit
timeExperiment [BrO3
–] , M [Br–] , M [H+] , M
1 0.0050 0.25 0.30 10
2 0.010 0.25 0.30 20
3 0.010 0.50 0.30 40
4 0.010 0.50 0.60 160
Initial Concentrations Rate in M per unit
timeExperiment [BrO3
–] , M [Br–] , M [H+] , M
1 0.0050 0.25 0.30 10
2 0.010 0.25 0.30 20
3 0.010 0.50 0.30 40
4 0.010 0.50 0.60 160In experiments A & B:• doubling [BrO3
–] doubles the rate.(other conc.’s kept constant)
• rate is 1st order w.r.t. [BrO3–]
Orders from Experimental Data: Rate varies with Concentration
rate = k [BrO3–]x
2 = [2]x
“w.r.t.” (with respect to)
1 = x
Initial Concentrations Rate in M per unit
timeExperiment [BrO3
–] , M [Br–] , M [H+] , M
1 0.0050 0.25 0.30 10
2 0.010 0.25 0.30 20
3 0.010 0.50 0.30 40
4 0.010 0.50 0.60 160In experiments B & C:• doubling [Br–] doubles the rate
(other conc.’s kept constant)• rate is 1st order w.r.t. [Br–]
Orders from Experimental Data: Rate varies with Concentration
rate = k [Br–]y
2 = [2]y
1 = y
Initial Concentrations Rate in M per unit
timeExperiment [BrO3
–] , M [Br–] , M [H+] , M
1 0.0050 0.25 0.30 10
2 0.010 0.25 0.30 20
3 0.010 0.50 0.30 40
4 0.010 0.50 0.60 160In experiments C & D:• doubling [H+] quadruples the rate
(other conc.’s kept constant)• rate is 2nd order w.r.t. [H+]
Orders from Experimental Data: Rate varies with Concentration
rate = k [H+]z
4 = [2]z
2 = z
Initial Concentrations Rate in M per unit
timeExperiment [BrO3
–] , M [Br–] , M [H+] , M
1 0.0050 0.25 0.30 10
2 0.010 0.25 0.30 20
3 0.010 0.50 0.30 40
4 0.010 0.50 0.60 160
Rate law is:rate = k [BrO3
–] [Br–] [H+]2
Orders from Experimental Data: Rate varies with Concentration
Orders found by experiment are:x = 1 , y = 1 , z = 2
1)…only found experimentally (from data).
2)…do NOT come from the coefficients of reactants of an overall reaction.
3)…represent the number of reactant particles (coefficients) in the RDS of the mechanism.
4)…zero order reactants have no effect on rate b/c they do not appear in the RDS of the mechanism (coefficient of 0 in RDS).
5)…typically 0, 1, 2, but can be any # or fraction
Orders in Rate Laws
OrderRate Law
k Units
M = ?∙M3
sM = ?∙M2
sM = ?∙M s
M = ? s
Units of k (rate constant)Units of k give info about order.
Rate is usually (M∙s–1), or (M∙min–1), etc.
M∙s–1 s–1 M–1∙s–1 M–2∙s–1
1 M2∙s
1M∙s
1 s
Ms
0th 1st 2nd 3rd
rate =k[A]0
rate =k[A]
rate =k[A]2
rate =k[A]2[B]
HW p. 618 #22,24,28
Determine the rate law for the reaction (from experimental data).
HW p. 619 #28
rate = k [ClO2]x [OH–]y
rate1
rate2
= (0.0248)= (0.00276)
= k (0.060)x(0.030)y
= k (0.020)x(0.030)y
0.060 x 0.020
0.030 y 0.030
(3)x (1)y
9 = (3)x
(0.0248) =(0.00276) =
9 =
x = 2
(a)
Determine the rate law for the reaction (from experimental data).
HW p. 619 #28
rate = k [ClO2]2 [OH–]y
rate3
rate2
= (0.00828)= (0.00276)
= k (0.020)2(0.090)y
= k (0.020)2(0.030)y
3 = (3)y y = 1
rate = k [ClO2]2 [OH–]1
OR rate = k [ClO2]2 [OH–]
(a)
Calculate the rate constant (with units).
HW p. 619 #28
rate = k [ClO2]2 [OH–]
Exp 1: (0.0248) = k (0.060)2(0.030)
(b)
(0.0248)(0.060)2(0.030)
k = 230
k =
M–2∙s–1
M = ?∙M2∙M s
Calculate the rate when [ClO2] = 0.010 M and [OH–] = 0.025 M.
HW p. 619 #28
rate = k [ClO2]2 [OH–]
rate = (230)(0.010)2(0.025)
(c)
rate = 0.00058 M∙s–1
Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.
Integrated rate laws express (reveal) the relationship between
concentration of reactants and time.
The differential rate law is usually just called “the rate law.”
Rate Laws
Integrated Rate Laws
ln[A]t
[A]0
= −kt
ln [A]t – ln [A]0 = –kt
ln [A]t = –kt + ln [A]0
y = mx + b
given on exam
Using calculus to integrate a first-order rate law gives us:
[A]0 = initial conc. of A at t = 0 .[A]t = conc. of A at any time, t .
If a reaction is first-order, a plot of ln [A] vs. t is a straight line, and the slope of the line will be –k.
First-Order Processes
ln [A]t = –kt + ln [A]0
y = mx + b
ln [A]
t
m = –k
first-order
First-Order Processes
CH3NC CH3CN
at 198.9 oC
First-Order Processes
• When ln P is plotted as a function of time, a straight line results.
• Therefore,The process is first-order.k is the negative slope: 5.1 10–5 s−1.
Second-Order Processes
Similarly, integrating the rate law for a process that is second-order in reactant A, we get…
1[A]t
= kt +1
[A]0
y = mx + b
1[A]t
–1
[A]0
= ktgiven
on exam
Second-Order Processes
1[A]t
= kt +1
[A]0
y = mx + b
If a reaction is second-order, a plot of 1/[A] vs. t is a straight line, and the slope of the line is k.
1[A]
t
m = k
second-order
Second-Order Processes
The decomposition of NO2 at 300°C is described by the equation
NO2 (g) NO (g) + 1/2 O2 (g)
and yields the following data:
Time (s) [NO2], M
0.0 0.01000
50.0 0.00787
100.0 0.00649
200.0 0.00481
300.0 0.00380
Second-Order Processes
• Graphing ln [NO2] vs. t yields:
Time (s) [NO2], M ln [NO2]
0.0 0.01000 −4.610
50.0 0.00787 −4.845
100.0 0.00649 −5.038
200.0 0.00481 −5.337
300.0 0.00380 −5.573
• The plot is NOT a straight line, so the process is NOT first-order in [A].
Second-Order Processes
• Graphing 1/[NO2] vs. t, however, gives this plot.
Time (s) [NO2], M 1/[NO2]
0.0 0.01000 100
50.0 0.00787 127
100.0 0.00649 154
200.0 0.00481 208
300.0 0.00380 263
• Because this IS a straight line, the process is second-order in [A].
If a reaction is zero-order, a plot of [A] vs. t is a straight line, with a slope = rate.
Zero-Order Processes
[A]
t
zero-order
zero-order
[A]
t
ln [A]t = –kt + ln [A]0
ln [A]
t
m = –kfirst-order
1[A]
t
m = ksecond-order
Summary ofIntegrated Rate Laws
andStraight-Line Graphs
1[A]t
= kt +1
[A]0
HW p. 620 #33, 38, 43
Half-life (t1/2):• time at which half of initial amount remains.
Half-life (t1/2) is constantfor 1st order only.
[A]t = 0.5 [A]0concentration
at time, t
initial concentration at time, t = 0
For a 1st order process:
0.5 [A]0
[A]0
ln = −kt1/2
ln 0.5 = −kt1/2
−0.693 = −kt1/2
= t1/20.693
k
ln [A]t – ln [A]0 = –kt
ln 0.5 [A]0 – ln [A]0 = –kt1/2
[A]t1/2 = 0.5 [A]0
given on
exam
Half-life (t1/2) depends on k:
A wooden object from an archeological site is subjected to radiometric dating by carbon-14.
The activity of the sample due to 14C is measured to be 11.6 disintegrations per second (current amount of C-14).
The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14).
The half-life of 14C is 5715 yr.
What is the age of the archeological sample?
Half-life & Radiometric Dating
A wooden object from an archeological site is subjected to radiometric dating by carbon-14.
The activity of the sample due to 14C is measured to be 11.6 disintegrations per second (current amount of C-14).
The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second (assumed as original amount of C-14).
The half-life of 14C is 5715 yr.
What is the age of the archeological sample?
Half-life & Radiometric Dating
t1/2 = 5715 yr
[A]0 = 15.2 initial dis/s at time, t0 = 0 s
[A]t = 11.6 current dis/s at time, t
t = ? yr
= 5715 yr 0.693
k
= k 0.693
5715 yr
First, determine the rate constant, k.
k = 1.21 10−4 yr−1
ln Nt − ln N0 = –kt
ln 0.5N0 − ln N0 = –k(5715)
0.5 N0
N0
ln = −k(5715)
–0.693–5715 = k
easy way
= t1/20.693
k
Half-life & Radiometric Dating
0.5ln = −k(5715)
–0.693 = −k(5715)
(given: t1/2 = 5715 yr)
Now we can determine t:
= −(1.21 10−4) t ln(11.6) – ln(15.2)
= −(1.21 10−4) t –0.270
= t 2230 yr
ln Nt − ln N0 = –kt
Half-life & Radiometric Dating
Summary 0th Order 1st Order 2nd OrderRate Law Rate = k Rate = k[A] Rate = k[A]2
Integrated Rate Law
[A] = –kt + [A]0 ln[A] = –kt + ln[A]0
Linear plot [A] ln[A]
k & slope of line
Slope = –k Slope = –k Slope = k
M∙s–1 s–1 M–1∙s–1
Half-Life depends on [A]0 depends on [A]0
0
1 1
[ ] [ ]kt
A A
1/ 2
0.693t
k
HW p. 620 #34,36,40,32
M = ?∙M2
sM = ?∙M
sM = ?
s
1[A]
Units of k
t t t
y = mx + b y = mx + b